$A$, $B$ have eigenvalue $\pi$ implies $A - B$ has eigenvalue $0$? I like to know if this is true or false and why.

If $A$ and $B$ are $n \times n$ matrices with the same eigenvalue $\pi$, then an eigenvalue of $A-B$ is $0$.

Kind regards,
Dieter.
 A: No, consider $A= \pmatrix{1 & 0 \\ 0 & -1}$ and $B= -A$. Then $A$ and $B$ have the same eigenvalues, but $A-B = \pmatrix{2&0 \\ 0 &-2}$ doesn't have  $0$  as an eigenvalue.
A: If I've interpreted the question correctly, the answer is no. Consider
$$A = \left[\begin{array}{cc}
2 & 0\\
0 & 1
\end{array}\right]\qquad\text{and}\qquad
B = \left[\begin{array}{cc}
1 & 0\\
0 & 2
\end{array}\right]$$
both of which have $2$ as an eigenvalue. Now consider $A - B$.
A: It will only work under the additional constraint, that they have a common eigenvector for that eigenvalue. Then you have $$Av = \lambda v = Bv \Rightarrow (A-B)v = Av-Bv = (\lambda - \lambda) v = 0$$
A: You got your counterexamples, but the "why" part might be more interesting. So, let us assume that we have matrices $A$ and $B$ with the common eigenvalue $\lambda$ (a bit more common than $\pi$). Then
$$Au = \lambda u, \quad Bv = \lambda v,$$
for some nonzero vectors $u$ and $v$. Now, if $u = v$, then
$$(A - B)u = Au - Bu = \lambda u - \lambda u = 0 = 0 u,$$
so zero is indeed an eigenvalue of $A-B$. However, if $u \ne v$, you cannot make such a conclusion. You can consider an eigenvector $w$ of $A - B$, but without any connection to $u$ and $v$, this is useless.
It probably seems obvious that the eigenvalues of $A - B$ are $\lambda_i - \mu_i$, where $\lambda_i$ are the eigenvalues of $A$, and $\mu_i$ are the eigenvalues of $B$, but this is far from correct, and works only if they have the same eigenvectors associated with them.
Also, note that the eigenvalues are never given in any specific order (i.e., there is no "first", "second",... eigenvalue), so you'd actually have $\lambda_i - \mu_j$ for all $i,j$, which is $n^2$ values if both $A$ and $B$ are diagonalizable, and $A - B$ can have at most $n$ eigenvalues, so $\lambda_i - \mu_j$ certainly cannot all be the eigenvalues of $A - B$.
